How do you convert $2=(3x+7y)^2-x$ into polar form?

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Answer 1 · 2017-12-17T17:28:46

$29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0$

$2=(3x+7y)^2-x$

$9x^2+42xy+49y^2-x=2$

After using $x=rcos(theta)$ and $y=rsin(theta)$ transormation,

$9r^2(cos(theta))^2+42r^2*cos(theta)*sin(theta)+49r^2(sin(theta))^2-rcos(theta)=2$

$9r^2*(1+cos(2theta))/2+21r^2*sin(2theta)+49r^2(1-cos(2theta))/2-rcos(theta)=2$

$29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0$

How do you convert 2=(3x+7y)^2-x2=(3x+7y)^2-x into polar form?